# Simplify Boolean Product of Sums Function

I've got a product of sums expression: F=(A'+B+C')&(A+D')(C+D')

I need to show it as a sum of products and then simplify it. Right now I got: F=(A'&D')+(A&B&C)+(B&D')+(C&D')

But the problem is that the values in the truth table are not the same. I believe that I've made some mistakes when trying to show it as a sum of products and simplyfing.

Thanks!

Your original expression is a product of sums:

(A' + B + C') (A + D') (C + D')


If you apply the and for the first two sums, you get:

(A'A + A'D' + AB + BD') (C + D')


A'A cancels out to false. In conjunction with the third sum, we get:

A'CD' + A'D'D' + ABC + ABD' + BCD' + BD'D'


Applying D'D' = D' gives us:

A'CD' + A'D' + ABC + ABD' + BCD' + BD'


A'CD' is covered by A'D' and can thus be omitted.

The minimized sum of products (the original six terms are covered by just four terms):

A'D' + C'D' + BD' + ABC


The terms of the expression shown in a Karnaugh-Veitch map: The diagram helps to visually grasp which term is covered by which larger term.